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Maximize light admitted through Norman window

  1. Jan 4, 2015 #1

    GreyNoise

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    1. The problem statement, all variables and given/known data
    A Norman window consists of a blue semicircular section surmounting a clear rectangular section as shown. The blue semicircular glass lets through half the light (per unit area) as the rectangular section glass (it is half as transparent or twice opaque). Find the radius 'r' of the window that admits the most light if the perimeter of the entire window is P

    2. Relevant equations
    I am using Ablue/Awhite = [(1/2)πr2] / [2hr] = πr / 4h where Ablue/Awhite are respective areas (please see attached for figure) and the perimeter P = 2h+2r+πr

    3. The attempt at a solution
    I used the perimeter P = 2h + 2r + πr to replace the factor 'h' in the ratio ( I mean I sub'd h = [P-2r-πr]/2 ) and then set d(Ablue/Awhite)/dr = 0 and solve. But the answer is

    r = 2P/(3π + 8)

    and I don't get anything close to that (I got r = P/4 among others). Can anyone suggest some hints to kick start this? I think I am just getting it wrong from the start.
     

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  2. jcsd
  3. Jan 4, 2015 #2

    Ray Vickson

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    You solve for ##h## in terms of ##r## from the perimeter constraint, then use that in some ratio or other. I do not understand why you do that. What is the relation of the ratio to the quantity you actually want to maximize?
     
  4. Jan 4, 2015 #3

    GreyNoise

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    Hello Ray V., thnx for the response. My reasoning was that if I minimized the blue area to the white area then the white area would have been maximized for some perimeter P and the maximum light would be admitted.
     
  5. Jan 4, 2015 #4

    haruspex

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    That's not a sound method, for a couple of reasons.
    Since the blue area admits some light through, it might not pay to maximise the clear area at the expense of the blue. You might find you have cut the blue area by a large amount in order to get a tiny increase in the clear area.
    Also, it is not immediately obvious that minimising the blue area while keeping the perimeter constant maximises the clear area.
     
  6. Jan 4, 2015 #5

    Ray Vickson

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    You are making the common error of "second-guessing" the solution. The problem asked for the maximum light through the whole window. Period.
     
  7. Jan 4, 2015 #6

    GreyNoise

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    The blue glass is twice as opaque per unit area as the clear, so can I leverage that information to get some relationship that I can optimize? I don't have a clue how I would do that. I have been through a dozen false starts for this one, and just can't seem to get it off the ground at all. I solved for a maximum total area for the window given some fixed P and got the answer
    [tex] r = P/(4+\pi)[/tex]
    (the answer given in the example). I did this to get some insight, but it did not help me for this one. I think am missing a basic concept here.
     
  8. Jan 4, 2015 #7

    SteamKing

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    The total amount of light depends directly on the area of the window. You initially were working with the ratio Ablue/Awhite, but that's not what gives you the amount of light passing thru the window. You are given that the blue window passes only 50% of the light for each square foot of area as the untinted window.

    What you want is to maximize the total amount of light coming thru the window as a whole, subject to the perimeter of the window being P units long. So, what expression would you write which gives the total amount of light coming thru the window?
     
  9. Jan 4, 2015 #8

    Ray Vickson

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    You know the area of the rectangular portion, and you know the area of the semi-circular portion (in terms of your variables, that is). The total light admitted is the light through the rectangle plus the light through the semicircle. Up to some arbitrary unit of light measurement, you can work out the total light admitted, just using the information supplied in the problem.
     
  10. Jan 4, 2015 #9

    haruspex

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    Let the dimensions of the rectangular area be 2r by y. Write down an expression for the perimeter and equate it to P. Write down another expression for the total light L coming through.
    If you differentiate both wrt r, remembering that y is a function of r so has some derivative y' = dy/dr, you will have altogether four equations. What do you think you should plug in for dL/dr? The remaining unknowns will be r, y, y', L. Solve.
     
  11. Jan 4, 2015 #10

    Ray Vickson

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    Suppose the light admitted per unit area of the white portion is ##k## (some constant that might be accessible from an engineering or optical handbook---but that does not really matter here). So what is the total light admitted through the whole white portion of the window? From the problem description, what is the light admitted per unit area through the blue portion (in terms of ##k##)? What is the light admitted through the entire blue portion of the window? What is the total light admitted through the whole window (both portions)?
     
  12. Jan 4, 2015 #11

    GreyNoise

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    Thnx for the input Ray V. If light through the blue glass is k (per unit area) then light through the clear glass is 2k (again per unit area). If the area of blue glass is [tex]\frac{1}{2}\pi r^2[/tex] then the light admitted through it is [tex]\frac{1}{2}\pi r^2 k[/tex] If the area for the white glass is [tex]2ry[/tex] then the light admitted through it is [tex]2ry*2k = 4ryk[/tex] The total light admitted is [tex]l_{tot} = l_{blue} + l_{clear} = \frac{1}{2}\pi r^2 k + 4ryk = (\frac{1}{2}\pi r^2 + 4ry)k[/tex] But the perimeter P is [tex]P = 2r + 2y + \pi r [/tex] so [tex]2y = P - 2r - \pi r[/tex] and substitution for y yields
    [tex]l_{tot} = (\frac{1}{2}\pi r^2 + 2rP - 4r^2 - 2\pi r^2)k[/tex] and finally applying d/dr gives
    [tex]\frac{d}{dr}l_{tot} = (\pi r + 2P - 8r - 4\pi r)k = 0[/tex]
    And that leads to (after a little simplification)
    [tex]r = \frac{2P}{8+4\pi-\pi} = \frac{2P}{8+3\pi}[/tex]
    Many many thnx for the help Ray V. I feel a little embarrassed to admit that I am teaching AP calc to high school juniors and seniors, and I wanted to present (or assign) this particular problem to them (they have solved the total area with me before) and I drew blanks on it (I do all homework from scratch before I assign it). As I tell my students, the solutions are always simple, but that is not be confused with obvious. Thnx again.
     
  13. Jan 5, 2015 #12

    Ray Vickson

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    Just a final point that is worth emphasizing to your students: we can, and should, erase the factor ##k## before solving the problem. The reason is that the admitted light function ##L(r,y)## has the form ##L(r,y) = k f(r,y)##, where ##f(r,y) = 4 r y + \frac{1}{2} \pi r^2##. Since ##k## is a fixed, positive constant, maximizing ##L## is equivalent to maximizing ##f##, in the sense that the same optimal ##(r,y)## applies to both problems. So, for example maximizing ##147 f## or ##0.25 f## or ##1,500,000 f## will all give the same ##(r,y)## solution as maximizing ##f##. I think you can see why, and it is worth getting the students to see it too. Basically, the use of ##k## acted as a formulation crutch that helped to clarify the underlying issues, but once it has served its purpose it can be dispensed with.

    BTW: there are some valuable on-line resources that may help with teaching formulations and solutions at the high-school level. For example, the website
    http://www.indiana.edu/~hmathmod/projects.html
    looks promising, although not all projects/models deal with calculus applications. The applications are mostly along the lines of "Operations Research" rather than mathematics per se.
     
    Last edited: Jan 5, 2015
  14. Jan 6, 2015 #13

    GreyNoise

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    Ray V.: Got it. Yes, I did point out to the students that we were using the factor k as a heuristsic device to assist us in this problem without losing generality. I had them eliminate it by noting that the optimizing diff equation was set to zero.

    (πr+2P−8r−4πr)k = 0

    So dividing both sides by k eliminated it whatever its value. Thnx also for the url source.

    Thnx haruspex and SteamKing for the hints (L = A*k).
     
  15. Jan 6, 2015 #14

    Ray Vickson

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    I think you are missing an important point---one that I was attempting to make, but perhaps not clearly enough: you can (and probably should) eliminate the ##k## factor before you start to optimize. So, you would not maximizing ##k (2rh + \pi r^2)## but just ##2rh + \pi r^2## itself! Of course the ##k## drops out of the solution to the optimization condition, but I was suggesting that you drop it before ever reaching that point.

    On the other hand, there are no rules about that, so if you prefer to keep the ##k## throughout you are perfectly free to do so without committing any errors. It just seems to be more-or-less standard practice to make such eliminations in applied optimization modelling---although I do not have specific references. Back in the Stone Age when I was teaching this stuff to Operations Research classes, I always tried to emphasize simplifying the formulation as much as allowable, so would always recommend moves like eliminating ##k## (with explanations of validity, of course). However, I never took marks off a student's work if they preferred not to adhere to those standards, unless they were told specifically to do so.
     
  16. Jan 6, 2015 #15

    haruspex

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    On the other hand, eliminating k late can be turned to didactic advantage. A lot of people posting their problems on this forum get stuck because there's some data item they've not been given (mass, often) and don't realise they can just plug in an unknown for it and see what happens.
     
  17. Jan 7, 2015 #16

    GreyNoise

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    Okay Ray V., I went back to your post (R.V. Monday 01/05 18:26) and reflected on it a bit. So in my 20/20 hindsight, I think your point was made (in my defense I am running on fumes this month for lack of sleep). But to be sure that I am on the same page, my take away from this is better or lesser style as opposed to rigor or correctness.
     
  18. Jan 8, 2015 #17

    Ray Vickson

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    Yes, to a large extent.
     
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